Divide the polynomials.
Solution: Usually, there are many different ways to divide polynomials. Here, we will use the method of polynomial long division. Notice the numerator is missing a $1^{\text{st}}$ degree term. Let's add it as $0x$. $\begin{array}{r} x+\phantom{1}3 \\ x-3|\overline{x^2+0x+\phantom{1}4} \\ \mathllap{-(}\underline{x^2-3x\phantom{+13}\rlap )} \\ 3x+\phantom{1}4 \\ \mathllap{-(}\underline{3x-\phantom{1}9\rlap )} \\ 13 \end{array}$ We get that the quotient is $x+3$ and the remainder is $13$, and therefore: $\dfrac{x^2+4}{x-3}=x+3+\dfrac{13}{x-3}$ [I want to see a different way of performing the division.]